# Volumes of Revolution

• Feb 25th 2009, 03:19 PM
terrytriangle
Volumes of Revolution
find the volume of the area enclosed by Y=X^2+1, Y=0, X=2 and revolved around X=3
Using the Disk Method

I used the shell method and got: 175pi/3

But I need to know how to set up the disk integral

Also,

Y=X^2+1, X=0, Y=5 revolved around Y=7

using the shell method i think i got 12pi, but that may be wrong, and I need to know how to set up the disk method.

thanks so much
• Feb 25th 2009, 10:58 PM
Volume of revolution
Hello terrytriangle
Quote:

Originally Posted by terrytriangle
find the volume of the area enclosed by Y=X^2+1, Y=0, X=2 and revolved around X=3

Please clarify this question. The curve $\displaystyle y = x^2 +1$, the line $\displaystyle y =0$ (i.e. the $\displaystyle x$-axis) and the line $\displaystyle x = 2$ does not enclose an area. Do you mean the area enclosed by

• $\displaystyle y=x^2 +1, y=0, x=2$ and $\displaystyle x=3$?
• Or $\displaystyle y=x^2+1, x=0, y=0$ and $\displaystyle x=2$?
• Or something else?

• Feb 25th 2009, 11:44 PM
earboth
Quote:

Originally Posted by terrytriangle
find the volume of the area enclosed by Y=X^2+1, Y=0, X=2 and revolved around X=3
Using the Disk Method

I used the shell method and got: 175pi/3

...

1. If I understand your question correctly the solid looks like a kind of funnel (upside down). (See attachment)

2. The solid consists of two different kinds of solids:
The base is a cylindrical ring with height of 1.
The top is the funnel mentioned above.

3. The base area of one disk is

$\displaystyle b = \pi((3-x)^2-1^2)=(x^2-6x+8)\cdot \pi$

the height of one disk is

$\displaystyle h = dy$ From

$\displaystyle \dfrac{dy}{dx} = 2x~\implies~dy=2x\cdot dx$

4. The complete volume is:

$\displaystyle V = \pi(3^2-1^2)\cdot 1+\int_0^2\left(\pi((x-3)^2-1^2)\cdot 2x \right)dx$

I leave the rest for you.

By the way: Your result must be too large: The solid can be placed completely into a cylinder with a base diameter of 6 and a height of 5. Such a cylinder has a volume of $\displaystyle 45\pi$
• Feb 26th 2009, 11:32 AM
terrytriangle
thank you i figured it out now
• Feb 26th 2009, 01:18 PM
HallsofIvy
earboth, terrytriangle said, "find the volume of the area enclosed by Y=X^2+1, Y=0, X=2 and revolved around X=3", which, as Granddad said, doesn't quite make sense. You have assumed the region enclosed by $\displaystyle y= x^2+ 1$, y= 0, x= 2, x= 3, revolved around x= 0.
• Feb 26th 2009, 10:57 PM
earboth
Quote:

Originally Posted by HallsofIvy
earboth, terrytriangle said, "find the volume of the area enclosed by Y=X^2+1, Y=0, X=2 and revolved around X=3", which, as Granddad said, doesn't quite make sense. You have assumed the region enclosed by $\displaystyle y= x^2+ 1$, y= 0, x= 2, x= 3, revolved around x= 0.